What determines the shape of a molecule in Chemistry?

The shape of a molecule is determined by the number of electron pairs around the central atom, including both bonding pairs and lone pairs. According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, electron pairs repel each other and arrange themselves as far apart as possible, which dictates the molecular geometry. This concept is crucial in the Cambridge International A Levels Chemistry curriculum, particularly in understanding Chemical Bonding.

How does the VSEPR theory help predict molecular shapes?

The VSEPR theory helps predict molecular shapes by considering the repulsion between electron pairs around a central atom. It states that electron pairs will arrange themselves to minimize repulsion, leading to specific geometric shapes. For example, a molecule with four bonding pairs like methane (CH4) will form a tetrahedral shape. This theory is a fundamental part of AS-Level Physical Chemistry under the Cambridge International A Levels syllabus.

What are some common molecular shapes and their bond angles?

Common molecular shapes include linear (180° bond angle), trigonal planar (120°), tetrahedral (109.5°), trigonal bipyramidal (90° and 120°), and octahedral (90°). These shapes result from different arrangements of electron pairs around the central atom. Understanding these shapes is essential for students studying Chemical Bonding in the Cambridge International A Levels Chemistry course.

Why do lone pairs affect the shape of a molecule differently than bonding pairs?

Lone pairs occupy more space around the central atom than bonding pairs because they are not shared between atoms. This increased repulsion can alter the bond angles and shape of the molecule. For instance, in water (H2O), the two lone pairs on oxygen push the hydrogen atoms closer together, resulting in a bent shape rather than a linear one. This concept is important for mastering Chemical Bonding in the AS-Level Physical Chemistry curriculum.

How can hybridization influence the shape of a molecule?

Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals, which can influence the shape of a molecule. For example, in methane (CH4), the carbon atom undergoes sp3 hybridization, resulting in a tetrahedral shape. Understanding hybridization is crucial for predicting molecular geometry and is a key topic in the Cambridge International A Levels Chemistry syllabus under Chemical Bonding.

Why is "Saying angles shrink 'because of lone pairs' without the repulsion order." a common mistake in Shapes of molecules?

Why it happens: Incomplete reasoning. How to avoid it: State lp–bp repulsion > bp–bp repulsion, which pushes bonding pairs together. Source: 9701 Examiner Reports 2022-2024

Why is "Counting a double bond as two regions (e.g. CO₂ as bent)." a common mistake in Shapes of molecules?

Why it happens: Counting bonds instead of regions. How to avoid it: A double/triple bond is ONE region; CO₂ is linear.

Why is "Naming NH₃ 'tetrahedral' because it has 4 regions." a common mistake in Shapes of molecules?

Why it happens: Confusing the electron arrangement with the molecular shape. How to avoid it: The SHAPE describes the atoms only: NH₃ is pyramidal (the lone pair isn't part of the named shape).

Why is "Giving 90° or 120° for a tetrahedral molecule." a common mistake in Shapes of molecules?

Why it happens: Mixing up the shape table. How to avoid it: Tetrahedral = 109.5°; learn each angle with its shape.

Chemical Bonding(AS-Level Physical Chemistry)Cambridge International A Levels Chemistry (9701)

Shapes of molecules

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Shapes of Molecules (VSEPR) — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Use VSEPR theory to predict molecular shapes and bond angles, account for the effect of lone pairs (NH₃ and H₂O), and extend the method to analogous molecules and ions.

What you’ll learn

Mapped to the Cambridge IGCSE 9701 syllabus (2025-2027).

3.5.1 — State and explain the shapes of, and bond angles in, molecules using VSEPR: BF₃ (trigonal planar 120°), CO₂ (linear 180°), CH₄ (tetrahedral 109.5°), NH₃ (pyramidal 107°), H₂O (non-linear 104.5°), SF₆ (octahedral 90°), PF₅ (trigonal bipyramidal 120° and 90°).
3.5.2 — Predict the shapes of, and bond angles in, molecules and ions analogous to those in 3.5.1.

The VSEPR principle

Electron pairs repel; they spread out to minimise repulsion. Lone pairs repel more than bonding pairs.

VSEPR = Valence Shell Electron Pair Repulsion. The idea:

Pairs of electrons around the central atom repel each other (like charges).
They arrange themselves to be as far apart as possible, which minimises repulsion and sets the shape.

Repulsion is not equal. A lone pair is held closer to the central atom and occupies more space, so: $\text{lone pair–lone pair} > \text{lone pair–bond pair} > \text{bond pair–bond pair}$

Method:

Count the regions of electron density around the central atom (bonding regions + lone pairs). A double or triple bond counts as one region.
Look up the basic geometry for that number of regions.
Adjust for lone pairs — they push bonding pairs closer together, reducing the bond angle.

Electron pairs repel and spread out maximally.
lp–lp > lp–bp > bp–bp repulsion.
Count regions (multiple bond = 1 region), then adjust for lone pairs.

The standard shapes and bond angles

Memorise these seven shapes with their angles and examples.

Regions	Lone pairs	Shape	Bond angle	Example
2	0	Linear	180°	CO₂, BeCl₂
3	0	Trigonal planar	120°	BF₃
4	0	Tetrahedral	109.5°	CH₄, NH₄⁺
4	1	Trigonal pyramidal	107°	NH₃
4	2	Non-linear (bent)	104.5°	H₂O
5	0	Trigonal bipyramidal	120° and 90°	PF₅
6	0	Octahedral	90°	SF₆

Note on CO₂. Each C=O double bond is one region → 2 regions → linear (180°), even though there are double bonds.

Linear 180°, trigonal planar 120°, tetrahedral 109.5°.
Trigonal bipyramidal 120°/90°, octahedral 90°.
CO₂ linear (2 regions despite double bonds).

The effect of lone pairs: NH₃ and H₂O

Same tetrahedral base as CH₄, but lone pairs squeeze the bond angle down.

CH₄, NH₃ and H₂O all have 4 regions of electron density (tetrahedral base), but different numbers of lone pairs:

CH₄ — 4 bonding pairs, 0 lone pairs → tetrahedral, 109.5°.
NH₃ — 3 bonding pairs, 1 lone pair → trigonal pyramidal, 107°. The lone pair repels the bonding pairs more strongly, squeezing them together.
H₂O — 2 bonding pairs, 2 lone pairs → non-linear (bent), 104.5°. Two lone pairs squeeze the angle even further.

Each lone pair reduces the bond angle by roughly 2.5° from the ideal 109.5°.

Same tetrahedral base; each lone pair (lp) repels more strongly and pushes the bond angle down.

CH₄ 109.5° (0 lp), NH₃ 107° (1 lp), H₂O 104.5° (2 lp).
Lone pairs repel more → smaller bond angle.
≈ −2.5° per lone pair.

Predicting shapes of other molecules and ions

Count regions and lone pairs on the central atom, then match to a known shape.

For any molecule or ion, count bonding regions + lone pairs on the central atom and match it to the table.

Worked examples:

NH₄⁺ — 4 bonding pairs, 0 lone pairs → tetrahedral, 109.5° (the lone pair of NH₃ is now used in the dative bond).
H₃O⁺ — 3 bonding pairs, 1 lone pair → trigonal pyramidal, ~107°.
BeCl₂ — 2 bonding pairs, 0 lone pairs → linear, 180°.
PCl₃ — 3 bonding pairs, 1 lone pair → trigonal pyramidal, ~107°.
SO₂ — 2 bonding regions + 1 lone pair → bent, ~120°.
CO₃²⁻ / NO₃⁻ — 3 bonding regions, 0 lone pairs → trigonal planar, 120°.

Tip. For ions, work out the dot-and-cross structure first (3.7) to count lone pairs correctly.

Count regions + lone pairs on the central atom → match the table.
NH₄⁺ tetrahedral; H₃O⁺ pyramidal; CO₃²⁻ trigonal planar.
Get lone pairs right via the dot-and-cross structure.

How it’s examined

Shapes and bond angles are guaranteed marks on Papers 1 and 2: name the shape, give the exact angle, and explain it with VSEPR. Examiner reports flag vague lone-pair explanations, giving 90°/120° for tetrahedral, and forgetting that NH₄⁺ is a perfect tetrahedron (109.5°).

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 3.5; 9701 Examiner Reports 2022-2024; 9701/12 & 9701/22 May/Jun 2024 question papers and mark schemes. Last reviewed 2026-05-20.

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Step-by-step worked examples — Shapes of molecules

Step-by-step solutions to past-paper-style questions on shapes of molecules, written exactly the way a tutor would explain them at the board.

1Shape and angle of CH₄ (2 marks)
Getting started• VSEPR
Question
State the shape and bond angle of a methane molecule, CH₄.
Step-by-step solution
1. Step 1
  4 bonding pairs, 0 lone pairs → 4 regions repelling equally.
2. Step 2
  Maximum separation of 4 regions is tetrahedral.
  $\text{tetrahedral},\ 109.5^\circ$
Answer
Tetrahedral, 109.5°.
Examiner tip
With no lone pairs the angle is the 'ideal' 109.5° — no reduction needed.
2Shape and angle of CO₂ (2 marks)
Getting started• VSEPR, double bond
Question
State the shape and bond angle of carbon dioxide, O=C=O.
Step-by-step solution
1. Step 1
  Each double bond counts as ONE region → carbon has 2 regions, no lone pairs.
2. Step 2
  Two regions point opposite → linear.
  $\text{linear},\ 180^\circ$
Answer
Linear, 180°.
Examiner tip
A double bond is a single region of electron density — CO₂ is linear, not bent.
3Shape and angle of NH₃ (3 marks)
Building confidence• VSEPR, lone pair
Question
State and explain the shape and bond angle of an ammonia molecule.
Step-by-step solution
1. Step 1
  Count regions: 3 bonding pairs + 1 lone pair = 4 regions (tetrahedral base).
2. Step 2
  Lone pair repels more than bonding pairs (lp–bp > bp–bp), squeezing the N–H bonds together.
3. Step 3
  Result: trigonal pyramidal, bond angle 107°.
Answer
Trigonal pyramidal, 107° (one lone pair on N).
Examiner tip
The angle mark needs the lone-pair-repulsion reasoning, not just '107°'.
4Predicting shapes of ions (3 marks)
Building confidence• VSEPR, ions
Question
Predict the shape and bond angle of (a) NH₄⁺ and (b) H₃O⁺.
Step-by-step solution
1. Step 1
  NH₄⁺: 4 bonding pairs, 0 lone pairs → tetrahedral, 109.5°.
2. Step 2
  H₃O⁺: 3 bonding pairs, 1 lone pair → trigonal pyramidal, ~107°.
Answer
NH₄⁺: tetrahedral, 109.5°. H₃O⁺: trigonal pyramidal, ~107°.
Examiner tip
For ions, adjust the electron count for the charge, then count bonding pairs + lone pairs as usual.
5Comparing bond angles, H₂O vs NH₃ (3 marks)
Stretch• VSEPR, lone pair
Question
Explain why the bond angle in H₂O (104.5°) is smaller than in NH₃ (107°).
Step-by-step solution
1. Step 1
  H₂O has two lone pairs; NH₃ has one.
2. Step 2
  More lone pairs → more lone-pair repulsion, which pushes the bonding pairs closer together.
3. Step 3
  So the bond angle in water is reduced further (104.5° vs 107°).
Answer
Water has 2 lone pairs vs 1 in ammonia → greater lone-pair repulsion → smaller angle.
Examiner tip
Each lone pair lowers the angle by roughly 2.5° from the ideal 109.5°.
6Expanded octets: PCl₅ and SF₆ (4 marks)
Stretch• VSEPR, expanded octet
Question
Predict the shape and bond angle(s) of (a) PCl₅ and (b) SF₆.
Step-by-step solution
1. Step 1
  PCl₅: 5 bonding pairs, 0 lone pairs → trigonal bipyramidal.
  $\text{angles } 120^\circ\text{ (equatorial) and } 90^\circ\text{ (axial)}$
2. Step 2
  SF₆: 6 bonding pairs, 0 lone pairs → octahedral.
  $\text{angles } 90^\circ$
Answer
PCl₅: trigonal bipyramidal (120° and 90°). SF₆: octahedral (90°).
Examiner tip
Period 3 central atoms can hold more than 8 electrons (expanded octet), giving 5- and 6-region shapes.

Key Formulae — Shapes of molecules

The formulae you need to memorise for shapes of molecules on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Electron regions (steric number)
$\text{regions} = (\text{atoms bonded}) + (\text{lone pairs})$
$atoms bonded$
number of σ-bonded atoms (a double/triple bond counts once)
$lone pairs$
lone pairs on the central atom
When to use
Set the basic geometry: 2 = linear, 3 = trigonal planar, 4 = tetrahedral, 5 = trigonal bipyramidal, 6 = octahedral.
Bond-angle reduction by lone pairs
$\text{angle} \approx \text{ideal} - 2.5^\circ \times (\text{lone pairs})$
$ideal$
the lone-pair-free angle for that geometry (e.g. 109.5°)
$lone pairs$
number of lone pairs on the central atom
When to use
Estimating the reduced angle: NH₃ 107° (1 lp), H₂O 104.5° (2 lp).

Key Definitions and Keywords — Shapes of molecules

Definitions to memorise and the exact keywords mark schemes credit for shapes of molecules answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

VSEPR theory
Examiner keyword
Valence Shell Electron Pair Repulsion: electron pairs around a central atom repel and arrange themselves as far apart as possible to minimise repulsion.
Lone pair
Examiner keyword
A pair of valence electrons not involved in bonding; repels more strongly than a bonding pair.
Repulsion order
Examiner keyword
lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair; this is why lone pairs shrink bond angles.
Region of electron density
Examiner keyword
A bonding region (single, double or triple bond — counted as one) or a lone pair, used to determine geometry.
Standard shapes and angles
Examiner keyword
Linear 180°; trigonal planar 120°; tetrahedral 109.5°; trigonal bipyramidal 120°/90°; octahedral 90°; (with lone pairs) pyramidal ~107°, bent ~104.5°.

Common Mistakes and Misconceptions — Shapes of molecules

The traps other students keep falling into on shapes of molecules questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Saying angles shrink 'because of lone pairs' without the repulsion order.
9701 Examiner Reports 2022-2024
Why it happens
Incomplete reasoning.
How to avoid it
State lp–bp repulsion > bp–bp repulsion, which pushes bonding pairs together.
✕Counting a double bond as two regions (e.g. CO₂ as bent).
Why it happens
Counting bonds instead of regions.
How to avoid it
A double/triple bond is ONE region; CO₂ is linear.
✕Naming NH₃ 'tetrahedral' because it has 4 regions.
Why it happens
Confusing the electron arrangement with the molecular shape.
How to avoid it
The SHAPE describes the atoms only: NH₃ is pyramidal (the lone pair isn't part of the named shape).
✕Giving 90° or 120° for a tetrahedral molecule.
Why it happens
Mixing up the shape table.
How to avoid it
Tetrahedral = 109.5°; learn each angle with its shape.

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Shapes of molecules

Short Study Notes in the form of Slides

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What you’ll learn

How it’s examined

1Shape and angle of CH₄ (2 marks)

2Shape and angle of CO₂ (2 marks)

3Shape and angle of NH₃ (3 marks)

4Predicting shapes of ions (3 marks)

5Comparing bond angles, H₂O vs NH₃ (3 marks)

6Expanded octets: PCl₅ and SF₆ (4 marks)

Electron regions (steric number)

Bond-angle reduction by lone pairs

VSEPR theory

Lone pair

Repulsion order

Region of electron density

Standard shapes and angles

✕Saying angles shrink 'because of lone pairs' without the repulsion order.

✕Counting a double bond as two regions (e.g. CO₂ as bent).

✕Naming NH₃ 'tetrahedral' because it has 4 regions.

✕Giving 90° or 120° for a tetrahedral molecule.

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Shapes of molecules — frequently asked questions